Question

If 25 ml of 1.50M LiOH are neutralized by 15ml of nitric acid what is the molarity of HNO3

Answer 1

LiOH + HNO3 --> H2O + LiNO3
OH- + H+  --> H2O
OH - = 1.5M * .025L = .0375 mol
.0375mol/.04L = 0.9375M
Since it's at the equivalence point, OH- = H+ concentration

[HNO3] = 0.9375M

Answer 2

Answer:

Ca = 2.5M

Explanation:

Volume of Acid (Va) = 15ml

Concentration of acid (Ca) = ?

Volume of base (Vb) = 25

Concentration of base (Cb) = 1.50M

The balanced chemical equation is given as;

LiOH + HNO3 --> LiNO3 + H2O

The formular relaing the parameters is given as;

(CaVa) / (CbVb) = Na/Nb

Where Na = Number of moles of Acid = 1

Nb = Number of moles of Base = 1

Making Ca subject of interest;

Ca = (Na * Cb * Vb) / (Nb * Va)

Ca = (1 * 1.50 * 25) / (1 * 15)

Ca = 37.5 / 15

Ca = 2.5M